1Spectrum Pricing Games with Spatial Reuse inCognitive Radio Networks

Gaurav S. Kasbekar and Saswati Sarkar

AbstractIn Cognitive Radio Networks (CRN), there aremultiple primary and secondary users in a region, and primariescan lease out their unused bandwidth to secondaries in exchangefor a fee. This gives rise to price competition among the primaries,wherein each primary tries to attract secondaries by settinga lower price for its bandwidth than the other primaries.Radio spectrum has the distinctive feature that transmissionsat neighboring locations on the same channel interfere witheach other, whereas the same channel can be used at far-offlocations without mutual interference. So in the above pricecompetition scenario in a CRN, each primary must jointly selecta set of mutually non-interfering locations within the region(which corresponds to an independent set in the conflict graphrepresenting the region) at which to offer bandwidth and theprice at each location. In this paper, we analyze this pricecompetition scenario as a game and seek a Nash Equilibrium(NE). We identify a class of conflict graphs, which we refer toas mean valid graphs, such that the conflict graphs of a largenumber of topologies that commonly arise in practice are meanvalid. We explicitly compute a symmetric NE in mean validgraphs and show that it is unique.

Index TermsCognitive Radio Networks, Price Competition,Game Theory, Nash Equilibrium, Spatial Reuse, Mean ValidGraphs

I. INTRODUCTION

Cognitive Radio Networks (CRN) [1] are emerging as apromising solution for the efficient use of spectrum. In thesenetworks, there are two types of spectrum users: (i) primaryusers who lease certain portions (channels or bands) of thespectrum directly from the regulator, and (ii) secondary userswho can use a channel when it is not used by the primary. Weconsider a CRN with multiple primary and secondary usersin a region. Time is slotted, and in every slot, each primaryhas unused bandwidth with some probability, which it wouldlike to lease to the secondaries. Now, secondaries would like tolease bandwidth from the primaries that offer it at a low price,which results in price competition among the primaries. Pricecompetition is naturally modeled using game theory [2], andhas been extensively studied in economics using, for example,the classical Bertrand game [5] and its variants.However, a CRN has several distinguishing features, which

makes the price competition very different from oligopoliesencountered in economics. First, in every slot, each primarymay or may not have unused bandwidth available. So aprimary who has unused bandwidth is uncertain about thenumber of primaries from whom it will face competition.

G. Kasbekar and S. Sarkar are with the Department of Electri-cal and Systems Engineering at University of Pennsylvania, Philadel-phia, PA, U.S.A. Their email addresses are kgaurav@seas.upenn.edu andswati@seas.upenn.edu respectively.Part of this paper was presented at MobiHoc10.

Setting a low price will result in unnecessarily low revenues inthe event that very few other primaries have unused bandwidth,because even with a higher price the primarys bandwidthwould have been bought, and vice versa. Second, spectrum isa commodity that allows spatial reuse: the same band can besimultaneously used at far-off locations without interference;on the other hand, simultaneous transmissions at neighboringlocations on the same band interfere with each other. Thus,spatial reuse provides an opportunity to primaries to increasetheir profit by selling the same band to secondaries at differentlocations, which they can utilize subject to satisfying theinterference constraints. So when multiple primaries ownbandwidth in a large region, each needs to decide on a set ofnon-interfering locations within the region, which correspondsto an independent set in the conflict graph representing theregion, at which to offer bandwidth. This is another sourceof strategic interaction among the primaries each primarywould like to select a maximum-sized independent set tooffer bandwith at; but if a lot of primaries offer bandwidthat the same locations, there is intense competition at thoselocations. So a primary would have benefited by insteadoffering bandwidth at a smaller independent set and charginghigh prices at those locations.Some progress has been made in addressing the issue of

bandwidth uncertainty, both in the CRN setting [19], [20] andin the context of price competition among multiple firms in theEconomics literature [8]. The issue of spatial reuse, however,arises specifically in the context of spectrum trading and hasnot been investigated either in prior work on price competitionin CRNs [11], [12], [13], [14], [15], [16] or otherwise. SeeSection II for a survey of related work.In this paper, we analyze price competition in CRNs jointly

considering both bandwidth uncertainty and spatial reuse, andspecifically focusing on the latter. We formulate the problemas a game in which each primary needs to select (i) a setof locations at which to offer bandwidth and (ii) the priceof bandwidth at each location. We seek to obtain a NashEquilibrium (NE) in this game. The challenge in doing sois that, since prices can take values from a continuous set,the strategy sets of primaries are uncountably infinite. So it isnot apriori clear whether a NE exists, and there is no standardalgorithm for finding a NE, unlike when each players strategyset is finite [2].We focus on symmetric NE, in which every player uses the

same strategy, since our game turns out to be a symmetricgame. Our first contribution is to prove a separation theorem(Section III-D), which states that in a symmetric NE, theprice distributions used by the primaries at different nodesare uniquely determined once the independent set selection

2distributions are obtained. We therefore focus on comput-ing the latter, which in turn provides the joint independentset and price selection strategies, by virtue of our priorresults [19], [20] that characterize price competition in theabsence of spatial reuse.

We focus on a class of conflict graphs that we refer to asmean valid graphs. These are graphs whose node set can bepartitioned into d disjoint maximal independent sets I1; : : : ; Id,for some integer d 2, and which satisfy another technicalcondition to be introduced later (in Section IV). As we showin Section VI-A, it turns out that the conflict graphs of a largenumber of topologies that arise in practice are mean valid. Inparticular, several lattice arrangements of nodes in two andthree dimensions are mean valid, e.g., a grid graph in twodimensions, such as that in part (b) of Fig. 3 or Fig. 13, whichmay be the conflict graph of shops in a shopping complex, theconflict graph of a cellular network with hexagonal cells (seeFigs. 7 and 8), a grid graph in three dimensions, which mayrepresent offices in a corporate building (see Fig. 6) etc.

We show that a mean valid graph has a unique symmetricNE; in this NE, each primary offers bandwidth only at someor all of the independent sets in I1; : : : ; Id with positiveprobability and with 0 probability at every other independentset. These probabilities (and thereby the NE strategies) canbe explicitly computed by solving a system of equations thatwe provide. The fact that primaries offer bandwidth with apositive probability at only a small number of independentsets is a surprising result, because in most graphs, includingthe examples in the previous paragraph, the number of in-dependent sets is exponential in the number of nodes. Ourcharacterization of the symmetric NE also reveals that whenthe probability q that a primary has unused bandwidth issmall, primaries only offer bandwidth at the larger independentsets out of I1; : : : ; Id and as q increases, primaries also startoffering bandwidth at the smaller ones. This is because, forgiven prices, a larger independent set yields a larger revenue.However, as q increases, the price competition at the largeindependent sets becomes intense and drives down the pricesand revenues at those independent sets. So primaries also offerbandwidth at the smaller independent sets.

The paper is organized as follows. We review related workin Section II and describe our model in Section III-A. InSection III-C, we provide a summary of the results on pricecompetition at a single location that were developed in ourprior work [19], [20]. We introduce mean valid graphs inSection IV and provide several examples. In Section V, weprove the theorem, discussed above, on characterization of theunique symmetric NE in mean valid graphs. In Section VI, weshow that the conflict graphs of several topologies of practicalinterest as well as some other common types of graphs aremean valid. In Section VII-A, we show that the mean validitycondition is a necessary condition for the existence of asymmetric NE of the above form when d = 2, and also findthe symmetric NE and prove its uniqueness in a specific nonmean valid graph.

II. RELATED WORK

Pricing related issues have been extensively studiedin the context of wired networks and the Internet;see [10] for an overview. Price competition among spec-trum providers in wireless networks has been studiedin [11], [12], [13], [14], [15], [16]. Specifically, Niyato et. al.analyze price competition among multiple primaries in CRNs[15], [16]. However, neither uncertain bandwidth availability,nor spatial reuse is modeled in any of the above papers. Also,most of these papers do not explicitly find a NE (excep-tions are [12], [15]). Our model incorporates both uncertainbandwidth availability and spatial reuse, which makes theproblem challenging; despite this, we are able to explicitlycompute a NE. Zhou et. al. [17] have designed double auctionbased spectrum trades in which an auctioneer chooses anallocation taking into account spatial reuse and bids. However,in the price competition model we consider, each primaryindependently sells bandwidth, and hence a central entity suchas an auctioneer is not required.In the economics literature, the Bertand game has been

traditionally used to study price competition in oligopolies [5]in this game, each seller quotes a price for a good, andthe buyers buy from the seller that quotes the lowest price.Several variants of the Bertrand game have also been studied,e.g., [6], [7], [8], [18]. Chawla et al. [18] consider pricecompetition in networks where each seller owns a capacity-constrained link, and decides the price for using it; the con-sumers choose paths they would use in the networks based onthe prices declared and pay the sellers accordingly. Note thatin our model, the sellers need to decide the locations at whichto offer bandwidth as well as the price at each location. Inaddition, the link capacities are deterministic in [18], whereasthe availability of bandwidth is random in our model. However,none of the above papers [6], [7], [8], [18] consider thespectrum-specific issue of spatial reuse, which introduces anew dimension, that each player not only needs to determinethe price of the commodity it owns (as in [6], [7], [8], [18]),but also select an independent set to compete in. The jointdecision problem significantly complicates the analysis.

III. MODEL, PROBLEM DEFINITION AND BACKGROUND

A. Model

Suppose there are n 2 primaries, each of whom owns achannel throughout a large region. Time is divided into slotsof equal duration. In every slot, each primary independentlyeither uses its channel throughout the region to satisfy its ownsubscriber demand, or does not use it anywhere in the region.A typical scenario where this happens is when primariesbroadcast the same signal over the entire region, e.g., if theyare television broadcasters. Let q 2 (0; 1) be the probabilitythat a primary does not use its channel in a slot (to satisfyits subscriber demand). For tractability, we assume that theprobability q is the same for all primaries; we discuss the effectof relaxing this assumption in Section IX. Now, the regioncontains smaller parts, which we refer to as locations. Forexample, the large region may be a state, and the locations maybe towns within it. We first assume that there are k secondaries

3at each location, and in Section VII-B outline how the resultscan be generalized to allow for random and potentially unequalnumber of secondaries at different locations.A primary who has unused bandwidth in a slot can lease

it out to secondaries at a subset of the locations, providedthis subset satisfies the spatial reuse constraints, which wedescribe next. The overall region can be represented by anundirected graph [4] G = (V;E), where V is the set of nodesand E is the set of edges, called the conflict graph, in whicheach node represents a location, and there is an edge betweentwo nodes iff transmissions at the corresponding locationsinterfere with each other. Recall that an independent set [4](I.S.) in a graph is a set of nodes such that there is no edgebetween any pair of nodes in the set. Now, a primary whois not using its channel must offer it at a set of mutuallynon-interfering locations, or equivalently, at an I.S. of nodes;otherwise secondaries1 will not be able to successfully transmitsimultaneously using the bandwidth they purchase, owing tomutual interference.A primary i who offers bandwidth at an I.S. I , must also

determine for each node v 2 I , the access fee, pi;v , to becharged to a secondary if the latter leases the bandwidth atnode v. A primary incurs a cost of c 0 per slot per nodefor leasing out bandwidth. This cost may arise, for example,if the secondary uses the primarys infrastructure to access theInternet.We assume that pi;v for each primary i and each node

v, for some constant > c. This upper bound may arise asfollows:1) The spectrum regulator may impose this upper bound

to ensure that primaries do not excessively overpricebandwidth even when competition is limited owing tobandwidth scarcity or high demands from secondaries,or when the primaries collude.

2) Alternatively, the valuation of each secondary for 1 unitof bandwidth may be , and no secondary will buybandwidth at a price that exceeds its valuation.

We assume that the primaries know this upper limit .Secondaries buy bandwidth from the primaries that offer

the lowest price. More precisely, in a given slot, let Z bethe number of primaries who offer unused bandwidth at anode. Then, since there are k secondaries at each node, thebandwidth of the min(Z; k) primaries that offer the lowestprices is bought (ties are resolved at random) at the node. Theutility of a primary i who offers bandwidth at an I.S. I andsets a price of pi;v at node v 2 I is given by

P(pi;v c),

where the summation is over the nodes v 2 I at which primaryis bandwidth is bought. (The utility is 0 if bandwidth is notbought at any node).Thus, each primary must jointly select an I.S. at which

to offer bandwidth, and the prices to set at the nodes in it.Both the I.S. and price selection may be random. Thus, astrategy, say i, of a primary i provides a probability massfunction (p.m.f.) for selection among the I.S. and the price

1Note that secondaries are usually customers or local providers, andpurchase bandwidth for communication (and not television broadcasts). Thus,two secondaries can not use the same band simultaneously at interferinglocations.

distribution it uses at each node (both selections contingenton having unused bandwidth). Note that we allow a primaryto use different (and arbitrary) price distributions for differentnodes (and therefore allow, but do not require, the selection ofdifferent prices at different nodes), and arbitrary p.m.f. (i.e.,discrete distributions) for selection among the different I.S.The case k n is trivial: in this case, the strategy of offering

bandwidth at a maximum-sized I.S. and setting the maximumprice at every node in the I.S. maximizes the utility of eachprimary i regardless of the strategies of the other primaries.This is because, since k n, the number of buyers at everynode is always greater than or equal to the number of sellers.So henceforth, we assume that k n 1.

B. Symmetric Nash Equilibrium

The vector ( 1; : : : ; n) of strategies of the pri-maries is called a strategy profile [5]. Let i =( 1; : : : ; i1; i+1; : : : ; n) denote the vector of strategiesof primaries other than i. Let Efui( i; i)g denote theexpected utility of primary i when it adopts strategy i andthe other primaries adopt i.Definition 1 (Nash Equilibrium (NE), Symmetric NE):

A Nash equilibrium (NE) is a strategy profile such thatno player can improve its expected utility by unilaterallydeviating from its strategy [5]. Thus, ( 1 ; : : : ;

n) is a NE if

for each primary i:

Efui( i ; i)g Efui( e i; i)g; 8 e i (1)A NE ( 1 ; : : : ;

n) is a symmetric NE if all players play

identical strategies under it, i.e., 1 = 2 = : : : =

n.

Equation (1) says that when players other than i play i, i maximizes is expected utility;

i is said to be its best

response [5] to i.Now, note that in the game defined in Section III-A, all the

n players (primaries) have identical strategy sets and utilityfunctions. Such a game is said to be a symmetric game. In asymmetric game, in practice, it is challenging to implement aNE that is not symmetric the simple example of two primariesand a NE of ( 1 ;

2) elucidates the inherent complications

in the current context. If 1 6= 2 , then since the gameis symmetric, ( 2 ;

1) also constitutes a NE. If player 1

knows that player 2 is playing 2 ( 1 respectively), it would

choose the best response 1 ( 2 respectively), but it can

not know player 2s choice between the two options withoutexplicitly coordinating with him, which is again ruled out dueto the competition between the two. Under symmetric NE,all players play the same strategy, and thus this quandary issomewhat limited symmetric NE has indeed been advocatedfor symmetric games by several game theorists [9]. The naturalquestion now is whether there exists at least one symmetricNE, and also whether there is a unique symmetric NE (onlyuniqueness will eliminate the above quandary). Note that somesymmetric games are known to have multiple symmetric NE,e.g. the Meeting in New York game (see [5], p. 221).We show the existence of a symmetric NE, by explicitly

computing one, and prove that it is the unique symmetric NEin our context.

4C. Single Location

In this subsection, we briefly summarize the main results forprice competition among multiple primaries and secondaries ata single location, which were dealt with in detail in [19], [20].Since there is only one location, there are no spatial reuse

constraints, and the strategy of a primary i is a distributionfunction (d.f.) i(:), which it uses to select the price.Let

w(q; n) =

n1Xi=k

n 1i

qi(1 q)n1i: (2)

Since each primary independently has unused bandwidth withprobability (w.p.) q, w(q; n) is the probability that k or moreout of n 1 primaries have unused bandwidth. We will lateruse the following fact [19]:Lemma 1: w(q; n) is a strictly increasing function of q for

fixed n.Now, let:

~p = w(q; n)( c): (3)In [19], we showed that in the price competition game at a

single location, there is a unique symmetric NE in which eachprimary randomizes over the prices in the range [~p; v] usinga continuous distribution function (d.f.) (:) (see Theorem 2in [19]). Also, (:) is strictly increasing on [~p; v]; in particular,every price in [~p; v] is a best response for each primary [19].An explicit expression for (:) has been computed in [19].Finally, under this symmetric NE, each primary receives anexpected payoff of [19]:

~p c = ( c)(1 w(q; n)) (4)

D. A Separation Result

Recall that a strategy of a primary consists of a p.m.f.over I.S. and price distributions at individual nodes. Wenow provide a separation framework from which the pricedistributions at individual nodes in a symmetric NE followonce the I.S. selection p.m.f.s are determined.Let I be the set of all I.S. in G. For convenience, we

assume that the empty I.S. I; 2 I and we allow a primary tooffer bandwith at I;, i.e. to not offer bandwidth at any node,with some probability. Consider a symmetric strategy profileunder which each primary offers bandwidth at I.S. I 2 I w.p.(I), where: X

I2I(I) = 1: (5)

The probability, say v , with which each primary offersbandwidth at a node v 2 V equals the sum of the probabilitiesassociated with all the I.S. that contain the node, i.e.

v =X

I2I :v2I(I) (6)

Now, considering that each primary has unused bandwidthw.p. q, it offers it at node v w.p. qv . The price selectionproblem at each node v is now equivalent to that for thesingle location case, the difference being that each primaryoffers unused bandwidth w.p. qv , instead of q, at node v.Thus:

Lemma 2: Suppose under a symmetric NE each primaryselects node v w.p. v if it has unused bandwidth. Then underthat NE the price distribution of each primary at node v is thed.f. (:) in Section III-C, with qv in place of q.Thus, a symmetric NE strategy is completely specified oncethe I.S. selection p.m.f. f(I) : I 2 I g (which will in turnprovide the vs via (6)) is obtained.

E. Node and I.S. Probabilities

Consider a symmetric NE where each primary uses thestrategy , under which it offers bandwidth at I.S. I 2 Iwith some probability (I). The probability, v , with whicheach primary offers bandwidth at a node v 2 V is determinedby the I.S. distribution f(I) : I 2 I g via (6).Now, for simplicity, we normalize c = 1. With w(q; n)

as in (2), let:

W () = (1 w(q; n))( c) = (1 w(q; n)): (7)By Lemma 2, and similar to (4) in the single location case,in a symmetric NE if primaries offer bandwidth at a nodewith probability (and play the single-node NE strategywith q in place of q at that node), then W () is themaximum expected payoff that each primary i can get at thatnode. It gets this payoff W () if it sets any price in therange [ w(q; n)( c); ] at that node. Under the abovesymmetric NE with strategy profile ( ; : : : ; ), each primaryoffers bandwidth at node v 2 V w.p. v . So the expectedpayoff of each primary i is given by:

Efui( ; i)g =Xv2V

vW (v): (8)

Now, in general, different I.S. distributions f(I) : I 2 I gcan result in the same node distribution 2 fv : v 2 V g.However, by (8), the expected payoff of each primary in asymmetric NE is completely determined by the node distri-bution, i.e. it is the same under different I.S. distributionsthat correspond to the same node distribution. So if primaryi knows the node distribution chosen by the other primaries,then it has sufficient information to choose its best response; itdoes not need to know their I.S. distribution in addition. Thus,the game aspect of the price competition, i.e. the strategicinteraction between the primaries, is completely determinedby the node distribution.We now introduce a definition:Definition 2 (Valid Distribution): An assignment fv : v 2

V g of probabilities to the nodes is said to be a valid distribu-tion if there exists a probability distribution f(I) : I 2 I gsuch that for each v 2 V , v =

PI2I :v2I (I).

Note that, given a valid distribution fv : v 2 V g, acorresponding I.S. distribution can be computed by solvingthe system of linear equations (6).Thus, we can equivalently define the strategy of a primary

in a symmetric NE as a node distribution fv : v 2 V g.So henceforth, we interchangeably speak of the strategy of aprimary as either an I.S. distribution f(I) : I 2 I g (note

2Although we refer to fv : v 2 V g as a distribution, note thatP

v2V vneed not equal 1 in general.

5that the price distribution follows by Lemma 2) or a nodedistribution fv : v 2 V g. Also, we say that the symmetricNE is unique if the node distribution fv : v 2 V g is unique.

IV. MEAN VALID GRAPHS

We now introduce a class of graphs, which we refer toas mean valid graphs. The motivation behind studying thesegraphs is that the conflict graphs of several topologies thatcommonly arise in practice are mean valid graphs. Also, as weshow in the next section, these graphs have a unique symmetricNE, which can be explicitly computed and has a simple form.

A. Definition

Definition 3 (Mean Valid Graph): We refer to a graph G =(V;E) as mean valid if it satisfies the following two condi-tions:1) Its vertex set can be partitioned into d disjoint maximal 3

I.S. for some integer d 2: V = I1[I2[: : :[Id, whereIj , j 2 f1; : : : ; dg, is a maximal I.S. and Ij \ Im = ;,j 6= m.Let jIj j =Mj ,

M1 M2 : : : Md; (9)and Ij = faj;l : l = 1; : : : ;Mjg.

2) For every valid distribution 4 in which a primary of-fers bandwidth at node aj;l w.p. j;l, j = 1; : : : ; d,l = 1; : : : ;Mj ,

dXj=1

j 1; (10)

where

j =

PMjl=1 j;lMj

; j 2 f1; : : : ; dg: (11)

We now explain the two conditions in Definition 3. Recallthat a graph G = (V;E) is said to be d-partite if V canbe partitioned into d disjoint I.S. I1; : : : ; Id [4]. For example,when d = 2, G is a bipartite graph. The first condition inDefinition 3 says that G is a d-partite graph and has theadditional property that each of I1; : : : ; Id is a maximal I.S.Now we explain Condition 2 in Definition 3. Let fj;l :

j = 1; : : : ; d; l = 1; : : : ;Mjg be an arbitrary valid distribution.Consider the distribution 0j;l = j , with j as in (11), i.e.for each j and l = 1; : : : ;Mj , 0j;l is set equal to the mean ofj;m;m = 1; : : : ;Mj . If (10) is true, then this distributionof means is a valid distribution because it corresponds tothe I.S. distribution f(Ij) = j ; j = 1; : : : ; d; (I;) =1 Pdj=1 j ;(I) = 0; I 6= I1; : : : ; Id; I;g. Thus, Condition2 in Definition 3 says that in G, the distribution of meanscorresponding to every valid distribution is valid. As we willsee in Section V, this condition is the crux behind the factthat in the symmetric NE in a mean valid graph, each primaryoffers bandwidth with equal probabilities at all the nodes inIj for every j = 1; : : : ; d.

3Recall that an I.S. I is said to be maximal if I [ fvg is not an I.S. forall v 2 V [4].

4Note that we write j;l in place of aj;l to simplify the notation.

B. Examples

Technical as Definition 3 may seem, it turns out that severalconflict graphs that commonly arise in practice are mean valid.For example, consider the following graphs:1) Let Gm denote a graph that is a linear arrangement of

m 2 nodes as shown in part (a) of Fig. 3, with an edgebetween each pair of adjacent nodes. As an example,this would be the conflict graph for locations along ahighway or a row of roadside shops.

2) We consider two types of m m grid graphs, denotedby Gm;m (see part (b) of Fig. 3) and Hm;m (seeFig. 13). In both these graphs, m2 nodes (locations) arearranged in a square grid. In Gm;m, there is an edgeonly between each pair of adjacent nodes in the samerow or column. In Hm;m, in addition to these edges,there are also edges between nodes that are neighborsalong a diagonal as shown in Fig. 13. For example,Gm;m or Hm;m may represent a shopping complex, withthe nodes corresponding to the locations of shops withWiFi Access Points (AP) for Internet access. Dependingon the proximity of the shops to each other and thetransmission ranges of the APs, the conflict graph couldbe Gm;m or Hm;m. Hm;m is also the conflict graph of acellular network with square cells as shown in Fig. 14.

3) Let Tm;m;m be a three-dimensional grid graph (seeFig. 6), which may, for example, be the conflict graphfor offices in a corporate building or rooms in a hotel.

4) The conflict graph (Fig. 8) of a cellular network withhexagonal cells (Fig. 7).

5) Consider a clique 5 of size e, where e 1 is any integer.This is the conflict graph for any set of e locations thatare close to each other.

All of the above are mean valid graphs:Theorem 1: The following graphs are mean valid, with d,

the number of disjoint maximal I.S., indicated in each case:1) a clique of size e 1 (d = e),2) a line graph Gm (d = 2),3) a two-dimensional grid graph Gm;m (d = 2),4) a two-dimensional grid graph Hm;m (d = 4),5) a three-dimensional grid graph Tm;m;m (d = 8).6) a cellular network with hexagonal cells, under Assump-

tion 1 in Section VI-A (d = 3).We defer the proof of Theorem 1 to Section VI-A. Also,

as we show in Section VI-B, some other common classes ofgraphs, such as a star and a -regular bipartite graph, are meanvalid as well.A graph obtained by considering the union of disjoint mean

valid graphs, all of which correspond to the same integer d,and then adding some edges to get a connected graph, is amean valid graph under some technical conditions 6, e.g., thecellular networks in a group of neighboring towns or the WiFinetworks in the departments of a university campus. Fig. 1illustrates the latter example.

5Recall that a clique or a complete graph of size e is a graph with e nodesand an edge between every pair of nodes [4].

6These technical conditions are stated in Lemmas 10 and 11 in Sec-tion VI-A.

6Fig. 1. The rectangles represent departments in a university campus andthe circles are the ranges of WiFi access points. The circles (nodes) in eachrectangle constitute a grid graph Hm;m, which is mean valid with d = 4 (seepart 4 of Theorem 1). The overall graph is also mean valid with d = 4. WithIj ; j 2 f1; 2; 3; 4g, being disjoint maximal I.S. as in Definition 3, the numberin each circle indicates the I.S. to which the corresponding node belongs, i.e.nodes corresponding to the circles numbered j 2 f1; 2; 3; 4g belong to I.S.Ij .

C. A Necessary and Sufficient Condition

We state a property of mean valid graphs for later use.Lemma 3: Let G = (V;E) be a graph that satisfies Condi-

tion 1 in Definition 3. Suppose I 2 I contains mj(I) nodesfrom Ij , j = 1; : : : ; d. G is mean valid if and only if:

dXj=1

mj(I)

Mj 1 8I 2 I (12)

V. SYMMETRIC NE IN MEAN VALID GRAPHS

In this section, we show that a mean valid graph has aunique symmetric NE; in this NE, in the notation of Def-inition 3, primaries offer bandwidth at all the nodes in Ij ,j 2 f1; : : : ; dg, with the same probability tj , i.e. j;l = tj8l = 1; : : : ;Mj , where ftj : j = 1; : : : ; dg is the uniquesolution of a set of equations that we provide.First, we will argue, by contradiction, that for each j,

j;l = j 8l = 1; : : : ;Mj , where j is given by (11). Inthe symmetric NE ( ; : : : ; ), by (7) and the discussion justafter it, primary 1 gets an expected payoff of W (j;l) at nodeaj;l; also, by (8), its total expected payoff is:

Efu1( ; 1)g =dX

j=1

MjXl=1

j;lW (j;l) (13)

Suppose j;l; l = 1; : : : ;Mj are not all equal for some j. By(7) and Lemma 1,W () is a strictly decreasing function of ;so primary 1 offers bandwidth with a high probability j;l atnodes aj;l at which it gets a low payoff W (j;l). This seemsto suggest that primary 1 could get a higher overall payoff byunilaterally switching to an alternative strategy, say 0, underwhich it decreases (respectively, increases) the node probabili-ties at nodes that yield a low (respectively, high) payoff, if sucha strategy 0 were to exist. This would contradict the fact that

the distribution fj;l : j = 1; : : : ; d; l = 1; : : : ;Mjg is primary1s best response and thereby imply that j;l; l = 1; : : : ;Mjmust be equal for every j = 1; : : : ; d.The existence of such a strategy 0 is guaranteed by

Condition 2 in Definition 3 that (10) holds for every validdistribution. Let 0 be a strategy under which primary 1 offersbandwidth at each node in Ij , j 2 f1; : : : ; dg w.p. j . Notethat

Pdj=1 j 1 by (10); so 0 is a valid distribution

since it corresponds to the I.S. distribution (Ij) = j ,j 2 f1; : : : ; dg, (I;) = 1

Pdj=1 j , (I) = 0, I 6=

I1; : : : ; Id; I;. By (8), the total expected payoff of primary1 if it plays strategy 0 is:

Efu1( 0; 1)g =dX

j=1

MjXl=1

jW (j;l) (14)

By (13) and (14):

Efu1( ; 1)g Efu1( 0; 1)g

=dX

j=1

0@MjXl=1

j;lW (j;l) j0@MjX

l=1

W (j;l)

1A1A(15)Now, we have the following algebraic fact, proved in theAppendix.Lemma 4: Let N 2 be an integer, 1; : : : ; N be real

numbers and =PN

i=1 iN . Let f(x) be any strictly decreasing

function of x. Then:

(NXi=1

if(i)) (NXi=1

f(i)) (16)

with equality iff 1 = : : : = N = .Intuitively, since f(:) is strictly decreasing, in the LHS of(16), the terms in which f(i) is large are multiplied by smallfactors i and vice-versa; on the other hand, all terms f(i)on the RHS are multiplied by the same factor . So the LHSis smaller.Now, as mentioned above, f() = W () = 1 w(q; n)

is a strictly decreasing function of . So by Lemma 4, theexpression in (15) is 0, with equality holding iff j;1 =: : : = j;Mj = j for each j 2 f1; : : : ; dg. But since is a best response, Efu1( ; 1)g Efu1( 0; 1)g. Sothe expression in (15) must equal 0 and hence j;1 = : : : =j;Mj = j for each j 2 f1; : : : ; dg.Now, suppose

Pdj=1 j < 1. Then primary 1 can unilat-

erally offer bandwidth at each node in Id with probability1Pd1j=1 j > d instead of d and increase its payoff. Thiscontradicts the fact that the distribution is a NE. So we musthave

Pdj=1 j = 1. Thus, we have shown the following:

Lemma 5: In a mean valid graph, under every symmetricNE, each primary offers bandwidth at each node in Ij w.p.tj , j 2 f1; : : : ; dg, for some tj 0, j = 1; : : : ; d, wherePd

j=1 tj = 1.A typical way in which the node probability distribution j;l =tj 8l = 1; : : : ;Mj , arises is via the I.S. distribution (Ij) =tj ; j = 1; : : : ; d;(I) = 0 8I 6= I1; : : : ; Id.The following result provides necessary conditions for a

distribution ftj : j = 1; : : : ; dg as in Lemma 5 to constitute asymmetric NE.

7Lemma 6: If a distribution ftj : j = 1; : : : ; dg as inLemma 5 constitutes a symmetric NE, then I1; : : : ; Id0 arebest responses and Id0+1; : : : ; Id are not, for some integerd0 2 f1; : : : ; dg. Also, each I 2 I containing a node from Ijfor some j > d0 is not a best response. Hence:

tj = 0; j > d0: (17)

Intuitively, a primary prefers to offer bandwidth at a largeI.S. because it gets some revenue at every node in the I.S.it selects and its total payoff is the sum of the revenues atthe nodes of the I.S. Also, recall that by (9), I1; : : : ; Id arein decreasing order of size. So a primary will (i) try to offerbandwidth only at the largest I.S. I1, (ii) offer bandwidth atthe next largest I.S. I2 as well with some probability only ifthe competition at I1 increases beyond a threshold, (iii) offerbandwidth at I3 as well with some probability only if thecompetition at I1 and I2 increases beyond a certain thresholdand so on. Hence, the set of best responses out of I1; : : : ; Idis of the form I1; I2; : : : ; Id0 for some 1 d0 d.Now, if primary i offers bandwidth at I.S. I 0 2 I , its overall

expected payoff, denoted by U1(I 0), is the sum of the expectedpayoffs at the nodes in I 0, which, by (7) and the discussionjust after it, is given by:

U1(I0) =

Xv2I0

W (v) =Xv2I0

(1 w(qv; n)): (18)

Now, consider a symmetric NE with ftj : j = 1; : : : ; dg as inLemma 5. By (18) and the fact that jIj j =Mj , the payoff ofprimary 1 if it offers bandwidth at Ij is:

U1(Ij) =MjW (tj) (19)

By Lemma 6, I1; : : : ; Id0 are best responses and Id0+1 isnot. So:

U1(I1) = : : : = U1(Id0) > U1(Id0+1)

Substituting (19) into the above and using (17) and the factthat W (0) = 1 w(0; n) = 1, we get:

M1W (t1) = : : : =Md0W (td0) > Md0+1 (20)

Thus, we have shown the following:Lemma 7: A distribution ftj : j = 1; : : : ; dg as in Lemma 5

that constitutes a symmetric NE must satisfy (17) and (20) forsome integer d0 2 f1; : : : ; dg.Lemma 7 provides necessary conditions for a distribution

ftj : j = 1; : : : ; dg to constitute a symmetric NE. Thefollowing lemma shows that these conditions are sufficient aswell.Lemma 8: Let 1 d0 d and t1; : : : ; td be a probability

distribution such that (17) and (20) hold. Then the symmetricstrategy profile in which every primary offers bandwidth ateach node in Ij w.p. tj , j 2 f1; : : : ; dg, is a NE.The proof of Lemma 8 (see the Appendix) is based on the

fact that the graph, being mean valid, satisfies Condition 2 inDefinition 3.The following technical lemma shows the existence and

uniqueness of a distribution (t1; : : : ; td) satisfying (17) and(20) for every value of q.

Lemma 9: For every q 2 (0; 1), there exists a unique integerd0 = d0(q) and a unique probability distribution (t1; : : : ; td)such that (17) and (20) hold. Also, d0(q) is an increasingfunction of q and, for every value of q, t1 t2 : : : td.Note that the fact that d0(q) is an increasing function of

q is consistent with the intuition that for small values of q,primaries tend to offer bandwidth at only the larger I.S. out ofI1; : : : ; Id, and as q, and thereby the competition from otherprimaries increases, they also choose the smaller ones. Also,the fact that t1 t2 : : : td for all q is consistent with theintuition that primaries offer bandwidth at the larger I.S. witha larger probability.Finally, putting together the above discussion, we get the

main result of this section:Theorem 2: In a mean valid graph, for every q 2 (0; 1),

there is a unique symmetric NE; in this NE, each primaryoffers bandwidth at every node in Ij , j 2 f1; : : : ; dg, w.p. tj ,i.e. j;l = tj , l = 1; : : : ;Mj , where (t1; : : : ; td) is the uniquedistribution satisfying (17) and (20).

Proof: By Lemma 5, under every symmetric NE, eachprimary must offer bandwidth at all the nodes in Ij ,j 2 f1; : : : ; dg, w.p. tj for some probability distribution(t1; : : : ; td). Also, by Lemma 7, (17) and (20) hold for thisdistribution. By Lemma 9, for a fixed value of q 2 (0; 1),there exists a unique distribution (t1; : : : ; td) satisfying (17)and (20). Finally, by Lemma 8, the strategy profile where eachprimary uses this distribution is a NE. The result follows.Thus, every mean valid graph has a unique symmetric NE,

which can be explicitly computed by solving the system ofequations (17) and (20).Example: Suppose there are n = 2 primaries and k = 1

secondary. Consider a grid graphHm;m, which was introducedin Section IV-B, with m = 7 (see Fig. 13). By part 4 ofTheorem 1, this is a mean valid graph and, in the notation ofDefinition 3, d = 4, the I.S. I1; I2; I3 and I4 are as describedin Section VI-A, and M1 = 16, M2 = M3 = 12, M4 = 9.The symmetric NE is of the form described in Theorem 2 withd0(q); t1; t2; t3 and t4 for different q 2 (0; 1) as follows:1) For 0 < q < 14 , d

0 = 1, t1 = 1, t2 = t3 = t4 = 0.2) For 14 q < 1516 , d0 = 3, t1 = 111

3 + 2q

, t2 = t3 =

111

4 1q

t4 = 0.

3) For 1516 q < 1, d0 = 4, t1 = 1499 + 13q

, t2 = t3 =

149

1q + 12

t4 =

149

16 15q

.

Note that, consistent with Theorem 2, d0(q) is an increasingfunction of q and t1 t2 t3 t4 for each value of q.In fact, for all q, t2 = t3, which is because I2 and I3 are ofthe same size. Fig. 2 plots t1; t2 and t4 versus q. For small q,primaries offer bandwidth at the largest I.S. I1 with probability1; but as q increases, the competition at I1 increases, inducingthe primaries to shift probability mass from I1 to the otherI.S. So t1 decreases in q. However, note that for all values ofq, t1 t2 t4 and t4 is very small (less than 0:02).Remark 1: The unique symmetric NE need not be pure

even with respect to the node selections, as the above exampleshows. However, this mixed choice is not really an artifact ofmixed price choice. For instance, consider a scenario where

80 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

q

t 1, t 2

, t 4

t1t2t4

Fig. 2. The figure shows the symmetric NE probabilities t1; t2 and t4 forthe example after Theorem 2.

all primaries must choose the same fixed price p0 (perhapsthe prices have been standardized because of governmentregulation). Suppose there are two nodes v1 and v2 connectedby an edge, two primaries (n = 2) and one secondary at eachnode (k = 1). Then it is easy to show that the strategy profileunder which each primary offers bandwidth at v1 and v2 w.p.1=2 each constitutes the unique symmetric NE.The intuition behind randomization across different I.S. in a

symmetric NE is that primaries would like to offer bandwidthat an I.S. at which other primaries do not offer bandwidth witha high probability, whereas in a symmetric NE that is pure withrespect to the node selection, all primaries offer bandwidth atthe same I.S.

VI. SOME SPECIFIC MEAN VALID GRAPHS

Theorem 2 provides the form of the symmetric NE in meanvalid graphs. So in this section, we identify some classes ofmean valid graphs.

A. Topologies that Commonly Arise in Practice

We now prove Theorem 1.The proof of part 1 of Theorem 1 is straightforward: let

fv1; : : : ; veg be the nodes of the clique. Ij = fvjg; j =1; : : : ; e are disjoint maximal I.S. whose union is V . Also,these are the only I.S. in the graph; so (12) holds and theclique is mean valid by Lemma 3.Next, we prove some lemmas that we use to prove the other

parts of Theorem 1.Lemma 10: Let G = (V;E) be a mean valid graph, where

V = I1 [ : : : [ Id and I1; : : : ; Id are disjoint maximal I.S.Let E0 E be any set such that no edge in E0 is betweentwo nodes in the same I.S. Ij , j 2 f1; : : : ; dg. Then the graphG0 = (V;E0) is mean valid.Thus, if a graph G is mean valid, then the graph G0 obtained

by adding edges in any fashion to G, while ensuring that Ij ,j = 1; : : : ; d continue to be I.S. in G0, is a mean valid graphas well.Lemma 11: Suppose for each i = 1; : : : ; N , Gi = (V i; Ei)

is a mean valid graph, where V i = Ii1 [ : : : [ Iid, Ii1; : : : ; Iidare disjoint maximal I.S., and jIij j = M ij , j = 1; : : : ; d. LetMi = (M i1; : : : ;M

id). If

Mi = ciM0; i = 1; : : : ; N (21)

for some vector M0 = (M01 ; : : : ;M0d ) and positive scalars

c1; : : : ; cN , then G = ([Ni=1V i;[Ni=1Ei) is mean valid.Lemma 11 says that if Gi; i = 1; : : : ; N are mean valid

graphs, then their union G is a mean valid graph as well pro-vided each of Gi; i = 1; : : : ; N contains (i) the same number,d, of disjoint maximal I.S., and (ii) the same proportion ofnodes in the d I.S. Ii1; : : : ; I

id. Since the union graph G is

a disconnected graph with N components, Lemma 11 is notuseful by itself to prove that a graph is mean valid. But it canbe effectively used in conjunction with Lemma 10 to combinea set of N mean valid graphs into a new connected meanvalid graph by (i) first considering their union, which is adisconnected graph, (ii) and then adding some edges to it tomake it connected.A useful special case is when each of these N graphs Gi is a

clique of size d (which is mean valid by Part 1 of Theorem 1)with vertex set V i = fvi1; : : : ; vidg. Note that these graphssatisfy the hypothesis of Lemma 11 with Iij = fvijg, M ij = 1,8i; j, M0 = (1; : : : ; 1) and ci = 1 8i. This special case canbe used to prove the mean validity of several of the graphsmentioned in Theorem 1, as we explain below.For an integer m 1, let me (respectively, mo) denote the

greatest even (respectively, odd) integer less than or equal tom.We now prove part 2 of Theorem 1. Consider a linear

graph Gm with node set fv1; v2; : : : ; vmg as shown in part(a) of Fig. 3. First, let m be even say m = 2N . Fori = 1; : : : ; N , let Gi be the clique of size 2 with the node setV i = fv2i1; v2ig and the edge between the two nodes. In thenotation of Lemma 11, let Ii1 = fv2i1g and Ii2 = fv2ig. ByLemma 11, G = G1[G2[: : :[GN is a mean valid graph withd = 2 and the disjoint maximal I.S. I1 = fv1; v3; v5; : : : vmogand I2 = fv2; v4; v6; : : : ; vmeg. We can obtain Gm by addingthe edges (v2; v3), (v4; v5), . . . , (v2N2; v2N1) to G asillustrated in part (a) of Fig. 4. Note that no edge is betweentwo nodes in the same I.S. Ij , j 2 f1; 2g; so the hypothesis ofLemma 10 is satisfied. Hence, Gm is mean valid by Lemma 10.The proof of the fact that Gm is also mean valid for m odd isdeferred to the Appendix.Now, we prove part 3 of Theorem 1. Consider Gm;m, where

m may be odd or even. Let vij be the node in the ith rowand jth column i; j 2 f1; : : : ;mg(see part (b) of Fig. 3). Westart with a line graph Gm2 , which is mean valid by part 2of Theorem 1, and add some edges to obtain Gm;m as shownin Fig. 5. Specifically, let Gm2 be the line graph with the setof nodes fv1;1, v1;2, . . . , v1;m, v2;m, v2;m1, . . . , v2;1, v3;1,v3;2, . . . , v3;m, v4;m, v4;m1, . . . g and an edge between eachpair of consecutive nodes in this order. Gm2 is mean validwith d = 2, and the disjoint maximal I.S. I1 = fv11, v13,. . . , v1;mo , v22, v24, . . . , v2;me , v31, v33, . . . , v3;mo , . . . g andI2 = fv12, v14, . . . , v1;me , v21, v23, . . . , v2;mo , v32, v34, . . . ,v3;me , . . . g. Gm;m can be obtained from Gm2 by adding theremaining edges shown dotted in Fig. 5. Note that no edge isbetween the same I.S. Ij ; j = 1; 2. So Gm;m is mean valid byLemma 10.Next, we prove part 4 of Theorem 1. Consider Hm;m (see

Fig. 13). As in Gm;m, let vij be the node in the ith row andjth column. Let d = 4, I1 = fv11, v13, v15, . . . , v1;mo , v31,

9Fig. 3. Part (a) shows a linear graph, Gm, with m = 8 and part (b) shows agrid graph, Gm;m, withm = 5. In both graphs, the darkened and un-darkenednodes constitute I1 and I2 respectively.

v33, v35, . . . , v3;mo , . . . g, I2 = fv12, v14, v16, . . . , v1;me ,v32, v34, v36, . . . , v3;me , . . . g, I3 = fv21, v23, v25, . . . , v2;mo ,v41, v43, v45, . . . , v4;mo , . . .g and I4 = fv22, v24, v26, . . . ,v2;me , v42, v44, v46, . . . , v4;me , . . . g (see part (b) of Fig. 4).Note that I1, I2, I3 and I4 are disjoint maximal I.S. For i; j 2f1; : : : ;m 1g, let Ci;j be the clique consisting of the nodesfvi;j ; vi;j+1; vi+1;j ; vi+1;j+1g and the edges among them (seeFig. 15). First, let m be even. The proof that Hm;m is meanvalid is similar to the above proof of mean validity of Gm withm even: we can obtain Hm;m by considering the union ofthe cliques Ci;j , i; j 2 f1; 3; 5; : : : ;m 1g, which is a meanvalid graph by Lemma 11, and then adding the remainingedges as illustrated in part (b) of Fig. 4. Note that no edge isbetween two nodes in the same I.S. Ij , j 2 f1; 2; 3; 4g; so thehypothesis of Lemma 10 is satisfied. Hence, Hm;m is meanvalid by Lemma 10. The proof of the fact that Hm;m is alsomean valid for m odd is deferred to the Appendix.The proof of part 5 of Theorem 1 is similar to that of

part 4: we outline the differences. For i; j; l 2 f1; : : : ;mg,let vijl be the node in the ith row, jth column and lthlevel (in the direction normal to the plane of the paper). Thenode set of Tm;m;m can be partitioned into 8 disjoint maximalI.S. I1; : : : ; I8 similar to I1; : : : ; I4 for Hm;m (see Fig. 6).Also, cliques Cijl, i; j; l 2 f1; : : : ;m 1g of size 8 eachcan be defined similar to the cliques Cij for Hm;m. For meven, we can obtain Tm;m;m by considering the union of thecliques Cijl, i; j; l 2 f1; 3; 5; : : : ;m 1g and then addingthe remaining edges. The fact that Tm;m;m is mean validthen follows from Lemmas 11 and 10. The proof of the factthat Tm;m;m is also mean valid for m odd is outlined in theAppendix.We now prove part 6 of Theorem 1. Consider a cellular

network as shown in Fig. 7, whose conflict graph is shown inFig. 8. The nodes in the graph can be partitioned into threedisjoint maximal I.S. I1, I2 and I3 as shown in Fig. 8. Weconsider this conflict graph with the following assumption,

Fig. 4. Part (a) (respectively, part (b)) shows the construction of G6(respectively, H4;4) from 3 (respectively, 4) cliques of size 2 (respectively,4) each. The solid edges constitute the cliques G1, G2, G3 (respectively,C1;1, C1;3, C3;1 and C3;3) and the dotted edges are those that are addedlater. The numbers next to the nodes shows the I.S. they are in, i.e., a nodelabeled j is in I.S. Ij , where j 2 f1; 2g (respectively, j 2 f1; 2; 3; 4g). Notethat no edge is between two nodes in the same I.S. Ij ; so the hypothesis ofLemma 10 is satisfied.

Fig. 5. The figure shows the construction of the grid graph Gm;m from theline graph Gm2 for m = 4. The solid edges constitute Gm2 and the dottededges are later added to obtain Gm;m. The un-darkened and darkened nodesconstitute I1 and I2 respectively in both Gm2 and Gm;m. Note that no edgeis between a node in I1 and a node in I2, so the hypothesis of Lemma 10 issatisfied.

which eliminates problems arising due to boundary effects.Assumption 1: There are an even number, say r, of rows of

nodes, each containing 3 nodes, for some integer 1.Under this assumption, as illustrated in Fig. 8, the graph canbe obtained by considering the union of r disjoint cliquesof size 3 each, which is a mean valid graph by Lemma 11,and then adding some edges. Note that no edge is betweentwo nodes in the same I.S. Ij ; j 2 f1; 2; 3g (see Fig. 8); sothe hypothesis of Lemma 10 is satisfied. Hence, the graph ismean valid by Lemma 10.Note that the above proof goes through if the graph can be

partitioned into cliques of size 3 even if Assumption 1 is notsatisfied. If the graph cannot be partitioned into cliques of size3, then the analysis is more complicated because of boundaryeffects. We omit this analysis for brevity.

B. Some Other Classes of Mean Valid Graphs

In this subsection, we show that some other common classesof graphs are mean valid. We focus on connected bipartitegraphs [4], which are of the form G = (V;E) where V =

10

Fig. 6. Part (a) shows a three-dimensional grid graph Tm;m;m for m = 5.It consists of periodic repetitions of the graph shown in part (b). Also, in part(b), the node labels show the I.S. I1; : : : ; I8 they are in, i.e. a node with thelabel j is part of the I.S. Ij , j 2 f1; : : : ; 8g.

Fig. 7. The figure shows a tiling of a plane with hexagons, e.g. cells in acellular network. Transmissions at neighboring cells interfere with each other.

Fig. 8. The figure shows the conflict graph of a hexagonal tiling of a plane.Both the solid and dotted edges are part of the graph. The nodes labelled j,j 2 f1; 2; 3g, are in I.S. Ij . There are four rows of nodes. The figure alsoshows the construction of the graph from cliques of size 3 each, shown by thesolid edges. The dotted edges are added later. Note that no edge is betweentwo nodes in the same I.S., so the hypothesis of Lemma 10 is satisfied.

A[B and every edge is between a node in A and a node in B.Without loss of generality, suppose jAj jBj. In the notationof Definition 3, d = 2, I1 = B and I2 = A. Also, a necessarycondition for a node distribution fi; i 2 A; j ; j 2 Bg, underwhich bandwidth is offered at node i 2 A (respectively, j 2 B)w.p. i (respectively, j), to be valid is that

i + j 1 8(i; j) 2 E: (22)This is because, if i+j > 1 for some (i; j) 2 E, then with apositive probability bandwidth would be offered at both nodesi and j, which are neighbors.Recall that a -regular graph is one in which the degree of

every node is [4].Proposition 1: A -regular bipartite graph is mean valid.Proof: Let jAj = N and jBj = M . First, we show that

N = M . Since edges are incident upon each node in A,

jEj = jAj = N. Similarly, jEj =M. So N =M .Now, let fi; i 2 A; j ; j 2 Bg be a valid distribution.

Adding (22) over all (i; j) 2 E, we get:X(i;j)2E

(i + j) jEj = N (23)

But since exactly edges are incident on each node:X(i;j)2E

(i + j) = (Xi2A

i +Xj2B

j) (24)

By (23) and (24),Pi2A iN

+

Pj2B jN

1So Condition 2 in Definition 3 is satisfied and the graph ismean valid.Recall that a star is a graph with a node a1 called the center,

nodes b1; : : : ; bM called the leaves, and edges (a1; bj); j =1; : : : ;M [4]. Note that this is a bipartite graph with edgesonly between the sets A = fa1g and B = fb1; : : : ; bMg.Proposition 2: A star is mean valid.Proof: Let f1; 1; : : : ; Mg be a valid distribution. By

(22),1 + j 1; j = 1; : : : ;M

Adding these M inequalities and dividing by M gives 1 +

1+:::+M

M 1.Now, note that every tree is a bipartite graph [4]. Given a

tree, suppose the root constitutes layer 1, the children of theroot constitute layer 2 and the children of all the nodes in layeri constitute layer i + 1, i = 2; 3; : : :. Not every tree is meanvalid; a counterexample is presented in Section VII-A (seeFig. 10). The following result provides a sufficient conditionfor a tree to be mean valid.Proposition 3: A tree in which every node in an odd layer

has exactly children is mean valid.Proof: Let a tree T in which every node in an odd layer

has children be given. Let Nj be the total number of nodesin the jth layer of T and N =

Pj oddNj .

Let Ai = faig, Bi = fbi;1; : : : ; bi;g and Gi be a star withcenter ai and leaves the nodes in Bi. Note that each Gi ismean valid by Proposition 2. Also, Gi; i = 1; : : : ; N , satisfythe hypothesis of Lemma 11 with d = 2, Ii1 = Bi, I

i2 = Ai,

M i1 = , Mi2 = 1 8i, M0 = (; 1) and ci = 1 8i. So by

Lemma 11, G = G1 [G2 [ : : : [GN is mean valid.Now, we will obtain T by adding some edges to G as

illustrated in Fig. 9. Let the center, a1, of G1 be the root of T .Note that its children are b1;1; : : : ; b1;, the leaves of G1. Forj 2 f1; : : : ; g, suppose b1;j has lj children. Join b1;j by anedge to each of the centers of lj stars out of G2; : : : ; GN , usinga different set of stars for each j. Thus, we have obtained thenodes in the first 4 layers of the tree and the edges connectingthem. Suppose a node b in layer 4 has l0 children. Join it tothe centers of l0 stars out of G1; : : : ; GN , which have not beenused so far. Proceed in this manner to get the tree T . Notethat there is no edge between two nodes in the same partitionof the tree, which is a bipartite graph (see Fig. 9); so thehypothesis of Lemma 10 is satisfied. Hence, T is mean validby Lemma 10.

11

Fig. 9. Part (a) shows a tree in which each node in an odd layer has exactly3 children. Part (b) shows the construction of the tree. We start with starswith 3 leaves each, whose edges are shown in bold and then add some edges,shown dotted. Note that none of the dotted edges is between two nodes inthe same partition of the bipartite graph, so the condition in Lemma 10 issatisfied.

VII. SOME ADDITIONAL TOPICS

A. Non Mean Valid Bipartite Graphs

We have shown in Theorem 2, that a mean valid graphhas a unique symmetric NE that has a simple form underthis NE, for every q 2 (0; 1), each primary offers bandwidthwith the same probability tj at all the nodes in each I.S. Ij ,j 2 f1; : : : ; dg. Thus, mean validity is a sufficient conditionfor an arbitrary graph to have a symmetric NE of the formin Theorem 2 for all values of q. The following result showsthat for connected bipartite graphs, mean validity is also anecessary condition.Theorem 3: Let G be a connected bipartite graph that is not

mean valid. If w(q; n) > 1M2M1 , then (I1) = t1, (I2) = t2,(I) = 0 8I 2 I ; I 6= I1; I2, is not a symmetric NE for anyvalue of t1 and t2.Now, we provide an example of a non mean valid bipartite

graph and find the symmetric NE and prove its uniqueness.The symmetric NE is not of the form in Theorem 2 for anyvalue of q 2 (0; 1).Let the set of nodes be A [ B, where A = fa1; a2; a3g

and B = fb1; b2; b3g, and let there be an edge between a1(respectively, b1) and every edge in B (respectively, A) (seeFig. 10). The only maximal I.S. are Iab = fa2; a3; b2; b3g,

Fig. 10. A non mean valid graph.

Ia = A and Ib = B. The I.S. Iab contains 2 nodes from eachof A and B, i.e., m1(Iab) = m2(Iab) = 2 in the notation ofLemma 3. Also, m1(Iab)3 +

m2(Iab)3 > 1; so (12) is not satisfied.

Hence, this is not a mean valid graph by Lemma 3.In every symmetric NE, (Ia) = ta, (Ib) = tb and

(Iab) = 1 ta tb for some 0 ta; tb 1.

Lemma 12: If w(q; n) > 12 , then f1(x) = 2W (1 x) W (x) has a unique root t1 2 [0; 1]. Also, 0 < t1 < 12 .The following theorem provides the symmetric NE in the

above graph for each value of q 2 (0; 1).Theorem 4: 1) If w(q; n) 12 , then the symmetric

strategy profile corresponding to ta = tb = 0 is theunique symmetric NE.

2) If w(q; n) > 12 , then the symmetric strategy profile inwhich ta = tb = t1, the root of f1(:), is the uniquesymmetric NE.

Note that Iab, which contains 4 nodes, is the largest I.S. Sofor all values of q, primaries offer bandwidth with positiveprobability at the I.S. Iab in the symmetric NE. Since Iabcontains nodes from both A and B, the node probabilities atdifferent nodes in A (and B) are different. Thus, the symmetricNE is not of the form in Theorem 2 for any value of q.Again, since Iab is the largest I.S., Theorem 4 shows,

consistent with intuition, that for small values of q, primariesoffer bandwith only at Iab with positive probability; whenq, and thereby the competition at Iab, increases beyond athreshold, they also offer bandwidth at A and B with positiveprobability.

B. Some Remarks

For simplicity, we have assumed that there are k secondariesat each node, where k is a constant. However, all our resultsreadily generalize to the case in which the number of secon-daries at node v 2 V is Kv , where fKv : v 2 V g are i.i.d.random variables such that the primaries apriori know only theprobability mass function (p.m.f.) for Kv , Pr(Kv = k) = k,but not the values of fKv : v 2 V g. Let:

w(q; n) =n1Xk=1

k

n1Xi=k

n 1i

qi(1 q)n1i (25)

All our results go through if w(q; n) in the above equation isused in place of (2).Now, in presence of spatial reuse, we would in general have

NE that are not symmetric. For example, suppose there are twonodes v1 and v2 connected by an edge, two primaries (n = 2)and one secondary at each node (k = 1). Then the strategyprofiles in which primary 1 offers bandwidth at node v1 andprimary 2 at node v2 w.p. 1, or vice versa, and both primariesset a price of w.p. 1, are NE, apart from the symmetric NEin Theorem 2. This is in contrast with the game at a singlelocation, in which the symmetric NE is the unique NE [20].

VIII. NUMERICAL STUDIESIn this section, we describe numerical computations that

are directed towards assessing the impact of price competitionamong the primaries on the aggregate revenue of the primariesand the affordability of spectrum for the secondaries. Towardsthat end, we compare the symmetric NE resulting from pricecompetition with a scheme, denoted by OPT, under which allthe primaries cooperate so as to maximize the sum of theirrevenues.We consider the specific case of a grid graph Hm;m, which

was introduced in Section IV-B (see Fig. 13). By part 4 of

12

Theorem 1, this is a mean valid graph and, in the notationof Definition 3, d = 4 and the I.S. I1; I2; I3 and I4 are asdescribed in Section VI-A. We use the parameter values n =10, k = 3, = 1; c = 0 and m = 7.In Hm;m, the symmetric NE is of the form in Theorem 2

and under OPT, the I.S. I1; : : : ; I4 are selected in order of sizeand all the primaries always select the highest price .Fig. 11 reveals, as expected, that price competition sig-

nificantly reduces the aggregate revenue of the primariesrelative to OPT. Also, overall, the ratio between the aggregaterevenues under the symmetric NE and under OPT decreasesas q increases since the competition increases. Fig. 12 showsthat under price competition, the expected price per unit ofbandwidth is lower at the nodes in the larger I.S. This isbecause primaries prefer larger I.S. and hence the competitionis more intense there, driving down the prices.

0 0.2 0.4 0.6 0.8 10

20

40

60

80

100

120

140

q

RN

E, R O

PT, Ef

ficie

ncy

(%

)

Efficiency ROPTRNE

Fig. 11. The figure plots the aggregate revenues of the primaries, RNE andROPT , under the symmetric NE and OPT respectively, and the efficiency ofthe symmetric NE, = RNE

ROPT(in %), versus q.

0 0.2 0.4 0.6 0.8 10.4

0.5

0.6

0.7

0.8

0.9

q

Mea

n Pr

ice

I4I2I1

Fig. 12. The figure shows the mean price of bandwidth, given that it isoffered, at a (fixed) node in each of I1; I2 and I4 vs q under the symmetricNE.

IX. DISCUSSION AND FUTURE WORK

For analytical tractability, we have considered the symmetriccase where each primary has unused bandwidth with the sameprobability q. Even for price competition at a single location,analysis of the asymmetric case, where the probabilities ofbandwidth availability of different primaries are not equal, hasbeen done only in some special cases, e.g., when there is onlyone secondary [20]. An important problem for future researchis the analysis of the asymmetric case in the spatial reusesetting.

APPENDIX

Let W () be as in (7). We will use the following resultthroughout.Lemma 13: (i) For 0 1, 0 W () < 1, (ii)

W (0) = 1, and (iii) W () is strictly decreasing in .Lemma 13 follows from (7), the fact that w(0; n) = 0 andLemma 1.

A. Proofs of results in Section IV

Proof of Lemma 3: Suppose G is mean valid. Fix anI 2 I . Let

1I(aj;l) =

1; if aj;l 2 I0; else

Consider a distribution fj;l : j = 1; : : : ; d; l = 1; : : : ;Mjgin which bandwidth is offered at node aj;l 2 Ij w.p. j;l =1I(aj;l). This is a valid distribution because it corresponds tothe distribution f(I) = 1; (I 0) = 08I 0 2 I ; I 0 6= Ig. Also,

MjXl=1

j;l =

MjXl=1

1I(aj;l) = mj(I); j = 1; : : : ; d (26)

Let j be given by (11). Since the graph is mean valid, (10)holds. Substituting

PMjl=1 j;l = mj(I) from (26) into (10),

we get (12).To prove the converse, suppose (12) holds. Let fj;l : j =

1; : : : ; d; l = 1; : : : ;Mjg be a valid distribution. By definition,there exists a distribution f(I) : I 2 I g such that:

j;l =X

I2I :aj;l2I(I) (27)

which can be written as:

j;l =XI2I

(I)1I(aj;l) (28)

Now,

dXj=1

PMjl=1 j;lMj

!

=dX

j=1

1

Mj

8

13

B. Proofs of results in Section V

The following lemma is used in the proof of Lemma 4.Lemma 14: Let N 2 be an integer and

1; : : : ; N ; f1; : : : ; fN be real numbers. Then:

N(NXi=1

ifi) (NXi=1

i)(NXi=1

fi) =X

1i d0,Mj = U1(Ij) < U. So the maximum payoffthat primary 1 can get at a node v 2 Ij , j > d0 is

1 Mj , define tj(x) = 0.As shown above, tj(Mj) = 0. So tj(x) is continuous on[M1W (1);M1]. Let,

T (x) =

dXj=1

tj(x) (38)

As shown above, h(tj) is strictly decreasing on 0 tj 1for j = 1; : : : ; d. So tj(x) = h1(x) is strictly decreasing forx Mj . Also, by definition, tj(x) = 0 on Mj < x M1. Soby (38), T (x) is strictly decreasing on [M1W (1);M1] (notethat t1(x) is strictly decreasing on x M1). Also, tj(M1) =0, j = 1; : : : ; d. So

T (M1) = 0: (39)

Now, for j = 1 and x = M1W (1), t1 = 1 satisfies (37). Sot1(M1W (1)) = 1 and hence, by (38):

T (M1W (1)) 1: (40)Now, since each tj(x), j = 1; : : : ; d, is continuous on

[M1W (1);M1], so is T (x) by (38). Hence, by (39), (40) andthe intermediate value theorem, the equation T (x) = 1 has asolution x 2 [M1W (1);M1], which is unique because T (x)is strictly decreasing. Let d0(q) = maxfj : Mj xg. Bydefinition of tj(x), for j = 1; : : : ; d0(q), MjW (tj(x)) = x

and for j > d0(q), Mj < x and hence tj(x) = 0.

Thus, (t1(x); : : : ; td(x)) satisfy (17) and (20). Also, by(38),

Pdj=1 tj(x

) = T (x) = 1; so (t1(x); : : : ; td(x)) is aprobability distribution. The result follows.Uniqueness: Fix q. We now show the uniqueness of d0(q)

and the distribution (t1; : : : ; td) satisfying (17) and (20).Assume, to reach a contradiction, that there exist e; f 2f1; : : : ; dg and probability distributions t = (t1; : : : ; td) ands = (s1; : : : ; sd) such that tj = 0 (respectively, sj = 0) forj > e (respectively, j > f ) and for some y and z:

y =M1W (t1) = : : : =MeW (te) > Me+1 (41)

z =M1W (s1) = : : : =MfW (sf ) > Mf+1 (42)

First, suppose e = f . If y = z, then by (41) and(42), MjW (tj) = MjW (sj), j = 1; : : : ; e. By part (iii)of Lemma 13, W (:) is a one-to-one function; so tj = sj ,j = 1; : : : ; e. Also, tj = sj = 0, j > e. So t = s.Now, suppose z > y. Then MjW (sj) > MjW (tj), j =

1; : : : ; e. So W (sj) > W (tj), and by part (iii) of Lemma 13,sj < tj , j = 1; : : : ; e. So 1 =

Pej=1 sj y is not possible. By symmetry,z < y is also not possible.Now, suppose e < f . Then by (41) and (42), z =

Me+1W (se+1) Me+1 < y. So for j 2 f1; : : : ; eg:MjW (sj) = z < y =MjW (tj)

which implies sj > tj . SoPe

j=1 sj >Pe

j=1 tj = 1, which isa contradiction. So e < f is not possible. By symmetry, e > fis also not possible. The result follows.Monotonicity Now, we show that d0(q) is an increasing

function of q. Suppose not. Then there exist q and q0 suchthat q < q0, d0(q) = e, d0(q0) = f and e > f . Hence, by (20)and (7), there exist probability distributions (t1; : : : ; td) and(s1; : : : ; sd) such that for some y and z:

y =M1(1 w(qt1; n)) = : : : =Me(1 w(qte; n)) > Me+1(43)

z =M1(1w(q0s1; n)) = : : : =Mf (1w(q0sf ; n)) > Mf+1(44)

Soy =Mf+1(1 w(qtf+1; n)) Mf+1 < z:

Hence, for j = 1; : : : ; f , Mj(1 w(qtj ; n)) < Mj(1 w(q0sj ; n)). So w(q0sj ; n) < w(qtj ; n). By Lemma 1, w(x; n)is strictly increasing in x. So q0sj < qtj . Since q < q0, tj > sj .Thus,

Pfj=1 tj >

Pfj=1 sj = 1, which contradicts the fact that

(t1; : : : ; td) is a probability distribution. The result follows.Finally, we show that t1 t2 : : : td. For 1 i < j

d0(q), MiW (ti) = MjW (tj) by (20). But Mi Mj by (9);so W (ti) W (tj) and hence, by part (iii) of Lemma 13,ti tj . For l > d0(q); tl = 0 by (17). The result follows.

C. Proofs of results in Section VI

Proof of Lemma 10: Since no edge in E0 is between twonodes in the same I.S. Ij , it follows that in G0, I1; : : : ; Id aredisjoint maximal I.S. whose union is V . Using the notationin Definition 3, let fj;l : j = 1; : : : ; d; l = 1; : : : ;Mjg be avalid distribution in G0. We will show that (10) holds. Thenit will follow from Definition 3 that G0 is mean valid.

15

Let IG0 (respectively, IG) be the set of I.S. in G0 (respec-tively, G). Since E E0, each I.S. in G0 is an I.S. in G aswell, i.e. IG0 IG.Now, since the distribution fj;lg is valid in G0, by def-

inition, there exists a distribution f0(I) : I 2 IG0g suchthat

v =X

I2IG0 :v2I0(I) 8v 2 V: (45)

Define a distribution on IG as follows:

(I) =

0(I) if I 2 IG00 if I 2 IG nIG0 (46)

By (45) and (46):

v =X

I2IG:v2I(I) 8v 2 V: (47)

So by definition, fi;jg is a valid distribution in G as well.Since G is mean valid, (10) holds, which completes the proof.

Proof of Lemma 11: First, note that f(I1j [: : :[INj ) : j =1; : : : ; dg are disjoint maximal I.S. in G; so the first conditionin Definition 3 is satisfied.Let fij;l : j = 1; : : : ; d; l = 1; : : : ;M ijg be a valid

distribution in Gi. Since Gi is mean valid:dX

j=1

0@PMijl=1 ij;lM ij

1A 1; i = 1; : : : ; N (48)Now, it is given that:

M ij = ciM0j ; i = 1; : : : ; N ; j = 1; : : : ; d (49)

Adding (49) over i = 1; : : : ; N :

M0j (c1 + : : :+ cN ) =M1j + : : :+M

Nj ; j = 1; : : : ; d (50)

Multiplying (48) by ci, using (49) and adding over i =1; : : : ; N , we get:

NXi=1

dXj=1

0@PMijl=1 ij;lM0j

1A c1 + : : :+ cNDividing both sides by c1 + : : :+ cN and using (50):

dXj=1

0@ PNi=1PMijl=1 ij;lM1j + : : :+M

Nj

1A 1So G satisfies the second condition in Definition 3 as well andhence is mean valid.

Proof of part 2 of Theorem 1: In Section VI-A, weshowed that Gm is mean valid for even m. Now, let m beodd, say m = 2N 1 for some integer N 2. Consider avalid distribution fi : i = 1; : : : ; 2N 1g, where i is theprobability with which bandwidth is offered at node vi. WithI1 and I2 as defined in Section VI-A, note that jI1j = N andjI2j = N 1. Let

1 =1 + 3 + : : :+ 2N1

N

and2 =

2 + 4 + : : :+ 2N2N 1

To show that Condition 2 in Definition 3 is satisfied, we needto show that 1 + 2 1, i.e.

(N 1)(1 + 3 + : : :+ 2N1)+ N(2 + 4 + : : :+ 2N2) N(N 1) (51)

Since G2N1 is a bipartite graph and the distribution fig isvalid, the necessary condition in (22) holds and in this casebecomes:

i + i+1 1; i = 1; 2; : : : ; 2N 2 (52)Now,

LHS of (51)= f(N 1)(1 + 2) + (2 + 3)g

+f(N 2)(3 + 4) + 2(4 + 5)g+f(N 3)(5 + 6) + 3(6 + 7)g+ : : :

+f2(2N5 + 2N4) + (N 2)(2N4 + 2N3)g+f(2N3 + 2N2) + (N 1)(2N2 + 2N1)g

f(N 1) + 1g+ f(N 2) + 2g+ : : :+f2 + (N 2)g+ f1 + (N 1)g (by (52))

= N(N 1)

which proves (51) and the result follows.Proof of part 4 of Theorem 1: In Section VI-A, we

showed that Hm;m is mean valid for even m. Now, let m beodd. With I1, I2, I3 and I4 as defined in Section VI-A, it iseasy to check that jI1j =

m+12

2, jI2j = m214 , jI3j = m

214

and jI4j =m12

2.Consider a valid distribution fz : z 2 V g, where z is the

probability with which bandwidth is offered at node z. Wenow show that the graph is mean valid by showing that (10)holds, which in this case becomes:

(m 1)2(Xz2I1

z) + (m2 1)(

Xz2I2

z) + (m2 1)(

Xz2I3

z)

+(m+ 1)2(Xz2I4

z) (m2 1)24

: (53)

Consider cliques Ci;j , i; j 2 f0; : : : ;mg. For i; j 2f1; : : : ;m 1g, Ci;j is as defined in Section VI-A. For ior j (or both) equal to 0 or m, let Ci;j be dummy cliques,defined for convenience (see Fig. 15). For i; j 2 f0; : : : ;mg:X

z2Cijz 1; (54)

because, if not, then bandwidth would be offered simultane-ously at two or more of the nodes in Cij (which are neighbors)with a positive probability. For i 2 f0; : : : ;mg, let:

ei =

m i; i oddi; i even (55)

For i; j 2 f0; : : : ;mg, letfij = eiej : (56)

16

Note that by definition of the cliques fCi;jg, node vijbelongs to each of the cliques Ci1;j1, Ci1;j , Ci;j1 andCi;j as shown in Fig. 16. So multiplying (54) by fij andadding over i; j 2 f0; 1; : : : ;mg gives:X

z2Vgzz g0 (57)

where,

gvij = fi1;j1 + fi1;j + fi;j1 + fij (58)

and

g0 =

mXi=0

mXj=0

fi;j =

mXi=0

mXj=0

eiej =

mXi=0

ei

!2

=

0@ mXi=0;i odd

(m i) +mX

i=0;i even

i

1A2 = (m2 1)24

(59)

We will show below that

gz =

8

17

Ci1;j1;l, Ci1;j;l1, Ci1;j;l, Ci;j1;l1, Ci;j1;l, Ci;j;l1,and Ci;j;l. Using this fact, gvijl for vijl in each of I1; : : : ; I8can be computed similar to the derivation of (60). Also, g0can be calculated similar to (59). Substituting these values offgz : z 2 V g and g0 into (57), we get (10) for Tm;m;m andthereby the mean validity follows from Definition 3.

D. Proofs of results in Section VII-AProof of Theorem 3: Suppose (I1) = t1, (I2) = t2,

where t1 + t2 = 1 is a symmetric NE. By (18):

U1(Ij) =MjW (tj); j = 1; 2: (62)

First, suppose t1 = 0; t2 = 1. Since (I2) = t2 > 0,I2 is a best response. By (62) and part (ii) of Lemma 13,U1(I1) = M1. Again, by (62), and since W (1) < 1 by part(i) of Lemma 13:

U1(I2) =M2W (1) < M2 M1 = U1(I1)which contradicts the fact that I2 is a best response. So t1 =0; t2 = 1 is not a symmetric NE.Now, suppose t1 = 1; t2 = 0. Then I1 is a best response.

Similar to the previous paragraph, U1(I1) = M1W (1) andU1(I2) =M2. So by (7):

U1(I1) U1(I2) =M11 M2

M1 w(q; n)

< 0

since w(q; n) > 1 M2M1 . This contradicts the fact that I1 is abest response. Thus, t1 = 1; t2 = 0 is not a symmetric NE.Suppose 0 < t1; t2 < 1. Let I 2 I be such that:

m1(I)

M1+m2(I)

M2> 1; (63)

which exists by Lemma 3 since G is not mean valid. Since(I1); (I2) > 0, I1 and I2 are best responses. So U1(I1) =U1(I2) = U

, where U is the maximum payoff of any I.S.By (62):

W (tj) =U

Mj; j = 1; 2 (64)

Now, by (18):

U1(I) = m1(I)W (t1) +m2(I)W (t2)

= Um1(I)

M1+m2(I)

M2

(by (64))

> U (by (63))

which contradicts the fact that U is the maximum payoff ofany I.S.

Proof of Lemma 12: By (2) and (7),W (:) is a continuousfunction. So f1(x) is continuous on [0; 1]. Also, w0(x; n) >0 8x 2 (0; 1) [19]. So by (7), W 0() < 0 8 2 (0; 1). Hence,for x 2 (0; 1):

f 01(x) = 2W 0(1 x)W 0(x) > 0:So f1(x) is strictly increasing on [0; 1] [3].Also, by (7), f1(0) = 2W (1)W (0) = 1 2w(q; n) < 0

since w(q; n) > 12 , and f112

=W

12

= 1w q2 ; n > 0.

So by the intermediate value theorem [3], f1(x) has a roott1 2

0; 12. Also, t1 is the unique root in [0; 1] since f1(x) is

strictly increasing.

1) Proof of Theorem 4: Consider a symmetric strategyprofile under which each primary offers bandwidth at Ia(respectively, Ib) w.p. ta (respectively, tb) and at Iab w.p.1 ta tb. By (6), the corresponding node probabilities area1 = ta, a2 = a3 = 1 tb, b1 = tb, b2 = b3 = 1 ta.So by (18), the total expected payoffs of primary 1 if it offersbandwidth at each of the three I.S. are:

U1(Iab) = 2W (1 tb) + 2W (1 ta) (65)U1(Ia) = 2W (1 tb) +W (ta) (66)U1(Ib) =W (tb) + 2W (1 ta) (67)

Intuitively, since Iab is the largest I.S., we expect that ina symmetric NE, primaries would not offer bandwidth at oneor both of Ia and Ib without offering it at Iab. The followingresult confirms this.Lemma 15: Let q 2 (0; 1) be arbitrary. None of the follow-

ing can hold in a symmetric NE: (i) ta = 1, (ii) tb = 1 (iii)0 < ta; tb < 1 and ta + tb = 1.

Proof: First, suppose ta = 1 in a symmetric NE. Sinceta > 0, Ia is a best response. Also, tb = 0. So by (65), (66),(7) and the fact that w(0; n) = 0:

U1(Iab) U1(Ia) = 1 + w(q; n) > 0So U1(Iab) > U1(Ia), which contradicts the fact that Ia isa best response. Thus, ta = 1 is not possible. By symmetry,tb = 1 is also not possible.Now, suppose 0 < ta; tb < 1 and ta+tb = 1. Since ta; tb >

0, Ia and Ib are best responses. So U1(Ia) = U1(Ib). By (66),(67), (7) and the fact that ta + tb = 1, we get w(qta; n) =w(qtb; n). So by Lemma 1, ta = tb = 12 . Hence, by (65), (66)and (7):

U1(Iab) U1(Ia) =1 w

q2; n

> 0

which contradicts the fact that Ia is a best response.Now we are ready to prove Theorem 4.Case 1: w(q; n) 12 . Let ta and tb be arbitrary. By (65),

(66) and (7):

U1(Iab) U1(Ia)= 1 2w(q(1 ta); n) + w(qta; n) 1 2w(q; n) (by Lemma 1) (68) 0

since w(q; n) 1

2

Note that if ta > 0, then the inequality in (68) is strict.So U1(Iab) > U1(Ia), which is a contradiction becauseta > 0 implies that Ia is a best response. Hence, ta = 0. Bysymmetry, tb = 0. If ta = tb = 0, then U1(Iab) U1(Ia)and U1(Iab) U1(Ib); so Iab is a best response, whichis consistent with the fact that it is played w.p. 1. Thus,ta = tb = 0 is the unique symmetric NE.Case 2: w(q; n) > 12 . By Lemma 15, ta + tb < 1 for every

symmetric NE and hence Iab is a best response. Now, supposeta = 0. By (65), (66), (7) and the fact that w(0; n) = 0, weget U1(Iab) U1(Ia) = 1 2w(q; n) < 0 since w(q; n) > 12 .So U1(Ia) > U1(Iab), which contradicts the fact that Iab is abest response. Hence, ta > 0. By symmetry, tb > 0.

18

Thus, all three of Ia, Ib and Iab are best responses. SoU1(Iab) = U1(Ia) = U1(Ib). Substituting (65), (66) and (67),these are satisfied iff ta = tb = t1, the root of f1(x) = 0. Thiscompletes the proof.

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